Find the reduced equation of the elypsis such that: The foci are $(0,6);(0,-6)$ and the larger axis has length $34$.
I did the following:
Taking the equation $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$, with $b^2=a^2-c^2$.
With the information given above, then the focal distance is $2c=12$ then $c=6.$ As the larger axis has length $34$, $a=17$, then $b^2=289-36=253$ and then $b=\sqrt{253}$.
The equation of the conic should be:
$$\cfrac{x^2}{289}+\cfrac{y^2}{253}=1$$
But it is:
$$\cfrac{x^2}{253}+\cfrac{y^2}{289}=1$$
According to the book answers. I understand that this conic has the numbers $a$ and $b$ shifted, I guess that it's because the larger axis is on the $y$-axis but I'm not sure if this is the rule to shift $a$ and $b$ because some time ago I did one exercise and this theory was invalid (assuming I got it correct). So, when shoul I actually shift $a$ and $b$?
The foci are (0,6);(0,-6). Hense, they are located on the y-asis. So, the larger axis is on y-axis and $b=17$.
As a consequence, $c^2=b^2-a^2$, but not $c^2=a^2-b^2$ as you did.
With $a^2=b^2-c^2$ you will obtain the expected result.